Preface

This secton provides a stream of examples demonstating applications of power series method for solving initial value problems for second order differential equations.

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Glossary

Complex exponents

Let us consider a homogeneous linear differential equation in the Frobenius form
\begin{equation} \label{EqComplex.1} x^2 y'' + xp(x)\, y' + q(x)\,y = 0 , \end{equation}
where p(x) and q(x) are holomorphic functions at the origin, which is a regular singular point for Eq.\eqref{EqComplex.1}. Here \( y' = {\text d}y/{\text d}x , \quad y'' = {\text d}^2 y/{\text d}x^2 \) are derivatives with respect to x, the independent avriable. The standard method of solution---Fuch--Frobenius method---consists in assuming a series solution of the form
\begin{equation} \label{EqComplex.2} y(x) = \sum_{n\ge 0} c_n x^{n+r} , \end{equation}
where the exponent r is chosen so that the leading coefficient c0 is non-zero, substituting into equation (1), and determining the coefficients ck recursively. The condition c0 ≠ 0 leads to the so-called indicial equation (a quadratic equation with real coefficients) whose roots determine r.

In this section, we discuss the general equation \eqref{EqComplex.1} where the roots of the indicial equation are complex.

Theorem 1: If the indicial equation corresponding to the series solution \eqref{EqComplex.2} of the differential equation \eqref{EqComplex.1} has complex roots, then the real and imaginary parts of Eq.\eqref{EqComplex.2} are two real independent solutions of Eq.\eqref{EqComplex.1} and hence constitute a basis for the solution space.
Suppose that y(x) = u(x) + jv(x) is a complex non-trivial solution of equation (1), where u(x) and v(x) are real non-zero functions. Substituting into equation (1), remembering that the coefficients of the equation are real functions, and setting real and imaginary parts equal to zero, we obtain
\[ x^2 u'' + xp(x)\, u' + q(x)\, u = 0 \qquad \mbox{and} \qquad x^2 v'' + xp(x)\, v' + q(x)\, v = 0 . \]
Hence, u(x) and v(x) are two real functions satisfying the differential equation. It remains to prove that they are linearly independent. Suppose, for example, that the root r = α + jβ (the other root is r = α − jβ) is used in generating the complex solution y(x). Since c0 ≠ 0 (the required condition for generating r), the leading term in the complex series solution for y(x) is
\[ c_0 x^r = c_0 x^{\alpha + {\bf j} \beta} = c_0 x^{\alpha} \left[ \cos \left( \beta \ln x\right) + {\bf j} \sin \left( \beta \ln x\right) \right] , \qquad {\bf j}^2 = -1. \]
Hence, the leading term in u(x) is
\[ c_0 x^{\alpha} \cos \left( \beta \ln x\right) \]
and that for v(x) is
\[ c_0 x^{\alpha} \sin \left( \beta \ln x\right) \]
Clearly u(x) and v(x) are linearly independent. ﹡ ⁎ ✱ ✲ ✳ ✺ ✻ ✼ ✽ ❋
This theorem suggests that the best and perhaps fastest way of generating the two real solutions is to use the Frobenius method to obtain just one complex solution using any one of the two complex conjugate roots, and then taking real and imaginary parts. It is also of interest to observe that, in a certain sense, the treatment of the complex case is unlike that of the real root case. In the real root case, the differential equation must be solved twice using the Frobenius method in order to obtain the two independent solutions. In the complex case, the differential equation is solved only once.

Let us consider a particular case when the coefficients of the differential equation \eqref{EqComplex.1} are two-term polynomials:
\begin{equation} \label{EqComplex.3} x^2 y'' + x\left( p_0 + p_1 x^m \right) y' + \left( q_0 + q_1 x^m \right) y = 0 , \end{equation}
Using the ansatz \eqref{EqComplex.2}, we obtain the series representations for its derivatives:
\begin{align*} y' &= \sum_{n\ge 0} \left( n+ r \right) c_n x^{n+r-1} , \\ y'' &= \sum_{n\ge 0} \left( n+ r \right) \left( n+ r -1 \right) c_n x^{n+r-2} . \end{align*}
Substituting into equation \eqref{EqComplex.3}, dividing out the common factor xr and adjusting some indices, we come to the relation
\[ \sum_{n\ge 0} \left[ \left( n+ r \right) \left( n+ r -1 \right) + p_0 \left( n+ r \right) + q_0 \right] c_n x^n + \sum_{n\ge m} \left[ p_1 \left( n0m+r \right) + q_1 \right] c_{n-m} x^n = 0. \]
Setting coefficients of like powers of x equal to zero, we obtain from the x0 coefficient:
\[ \left[ r \left( r-1 \right) + p_0 r + q_0 \right] c_0 = 0 , \]
from the x1 coefficient:
\[ \left[ r \left( r+1 \right) + p_0 \left( r+1 \right) + q_0 \right] c_1 = 0 , \]
⋮
from the xm-1 coefficient:
\[ \left[ \left( m-1+r \right) \left( m-2+r \right) + p_0 \left( m- 1 +r \right) + q_0 \right] c_{m-1} = 0 , \]
and the coefficient of xn, n ≥ m, leads to the recurrence relation
\[ c_n = \frac{p_1 \left( n-m+r \right) + q_1}{n^2 + n \left( p_0 -1 + 2r \right)} \, c_{n-m} , \qquad n=m, m+1, \ldots . \]
From the x0 coefficient and the condition that c0 ≠ 0, we get the indicial equation
\begin{equation} \label{EqComplex.4} r \left( r-1 \right) + p_0 r + q_0 = 0 , \end{equation}
whose complex roots are
\begin{equation} \label{EqComplex.5} \left\{ \begin{array}{c} r \\ \overline{r} \end{array} \right\} = \frac{1 - p_0}{2} \pm {\bf j}\,\frac{\sqrt{4q_0 -(p_0 -1)^2}}{2} \qquad ({\bf j}^2 =-1) , \end{equation}
where \( 4q_0 -(p_0 -1)^2 > 0. \)

All the coefficients ck, k = 1,2, …, m−1, equal zero since the coefficient of ck reduces, on using the indicial equation, to \( k^2 - 2k + p_0 k + 2kr , \) which is never zero since r has a non-zero imaginary part. Thus, with n = mp, the recurrence relation in the new index p reduces to

\begin{equation} \label{EqComplex.6} c_{mp} = - \frac{p_1 \left[ m(p-1) + r \right] + q_1}{m^2 p^2 + mp \left[ p_0 -1+2r \right]} \, c_{m(p-1)}, \qquad p = 1,2,\ldots . \end{equation}
Substituting the explicit expression for the root r in equation \eqref{EqComplex.6}, we have, in polar form,
\begin{equation*} %\label{EqComplex.7} c_{mp} = \sqrt{A_p^2 + B_p^2} \,\exp \left\{ {\bf j} a_p \left[ \arctan\left( \frac{| B_p |}{| A_p |} \right) + b_p \pi \right] \right\} c_{m(p-1)} \end{equation*}
where
\[ A_p = \frac{\left( p_1 /2 \right) \left[ 4q_0 - (p_0 -1)^2 \right] -mp \left[ p_0 m\left( p-1 \right) + q_1 - (p_1 /2) \left( p_0 -1 \right) \right]}{mp \left[ m^2 p^2 + 4q_0 - \left( p_0 -1 \right)^2 \right]} , \]
\[ B_p = \sqrt{4q_0 -(p_0 -1)^2} \,\frac{p_1 m \left( p-1 \right) + q_1 -(p_1 /2)\left( p_0 -1 \right) -(mp/2) \,\sqrt{4q_0 -(p_0 -1)^2}}{mp \left[ m^2 p^2 + 4q_0 - \left( p_0 -1 \right)^2 \right]} , \]
\[ \left. \begin{array}{lll} a_p = \phantom{-}1 & b_p = \phantom{-}0 & \mbox{when } A_p > 0, \ B_p > 0, \\ a_p = \phantom{-}1 & b_p = \phantom{-}1 & \mbox{when } A_p < 0, \ B_p < 0, \\ a_p = -1 & b_p = \phantom{-}0 & \mbox{when } A_p > 0, \ B_p < 0, \\ a_p = -1 & b_p = -1 & \mbox{when } A_p < 0, \ B_p > 0, \end{array} \right\} \]
and
\[ 0 \le \arctan \left( \frac{| B_p |}{| A_p |} \right) \le \frac{\pi}{2} . \]
Using the recurrence relation \eqref{EqComplex.6}, and recursively substituting back for
\[ c_{m(p-1)}, \ c_{m(p-2)}, \ \ldots , \ \]
c_{m1}, we obtain
\begin{equation} \label{EqComplex.7} c_{mp} = \prod_{k=1}^p \sqrt{A_p^2 + B_p^2} \,\exp \left\{ {\bf j} \sum_{j=1}^p a_j \left[ \arctan\left( \frac{| B_j |}{| A_j |} \right) + b_j \pi \right] \right\} c_{m0} . \end{equation}
Letting cm0 = 1, and remembering that \( \displaystyle x^{{\bf j}\beta} = \exp \left\{ {\bf j}\beta \ln x \right\} , \) the solution in complex form becomes
Example 1: Consider the differential equation
\[ x^2 y'' - x\,y' + \left( x + 2 \right) y = 0 . \]
In this case, m = 1, p0 = −1, p1 = 0, q0 = 2, and q1 = 2. The roots of the indicial equation are
\[ \left\{ \begin{array}{c} r \\ \overline{r} \end{array} \right\} = 1 \pm {\bf j} . \]
The coefficients are
\[ A_p = -\frac{1}{p^2 + 4} , \qquad B_p = \frac{2}{p \left( p^2 + 4 \right)} . \]
   ■
Example 2: Consider the differential equation
\[ x^2 y'' - x\,y' + \left( x^2 + 2 \right) y = 0 . \]
This differential equation belongs to the class of equations \eqref{EqComplex.3}, and has the same indicial equation.    ■

 

  1. Dettman, J.W., Power Series Solutions of Ordinary Differential Equations, The American Mathematical Monthly, 1967, Vol. 74, No. 3, pp. 428--430.
  2. Grigorieva, E., Methods of Solving Sequence and Series Problems, Birkhäuser; 1st ed. 2016.
  3. Neuringer, J., The Frobenius method for complex roots of the indicial equation, International Journal of Mathematical Education in Science and Technology, 1978, Vol. 9, No. 1, pp. 71--77. https://doi.org/10.1080/0020739780090110

 

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